Question

giải phương trình sau

a) 0,75x ( x + 5 ) = ( x + 5 ) ( 3 - 1,25x )

b) \(\frac{4}{5}\)- 3 = \(\frac{1}{5}\)x ( 4x - 15 )

c) ( x - 3 ) - \(\frac{\left(x-3\right)\left(2x-5\right)}{6}\)= \(\frac{\left(x-3\right)\left(3-x\right)}{4}\)

d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}\)+ 5( 3x + 1 ) = \(\frac{2\left(2x+1\right)\left(3x+1\right)}{3}\)+2x ( 3x +1 )

Answer 1

a) 0,75x(x + 5) = (x + 5)(3 - 1,25x)

<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = (x + 5)(3 - 1,25x) - (x + 5)(3 - 1,25x)

<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = 0

<=> (x + 5)(0,75 + 1,25x - 3) = 0

<=> (x + 5)(2x - 3) = 0

<=> x + 5 = 0 hoặc 2x - 3 = 0

<=> x = -5 hoặc x = 3/2

b) 4/5 - 3 = 1/5x(4x - 15)

<=> -11/5 = x(4x - 15)/5

<=> -11 = x(4x - 15)

<=> -11 = 4x² - 15x

<=> 11 + 4x² - 15x = 0 

<=> 4x² - 4x - 11x + 11 = 0

<=> 4x(x - 1) - 11(x - 1) = 0

<=> (4x - 11)(x - 1) = 0

<=> 4x - 11 = 0 hoặc x - 1 = 0

<=> x = 11/4 hoặc x = 1

c) \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

<=> 12x - 36 - 2(x - 3)(2x - 5) = 3(x - 3)(3 - x)

<=> 12x - 36 - 4x² + 10x + 12x - 30 = 9x - 3x² - 27 + 9x

<=> 34x - 66 - 4x² = 18x - 3x² - 27

<=> 34x - 66 - 4x² - 18x + 3x² + 27 = 0

<=> 16x - 39x - x² = 0

<=> x² - 16x + 39x = 0

<=> (x - 3)(x - 13) = 0

<=> x - 3 = 0 hoặc x - 13 = 0

<=> x = 3 hoặc x = 13

d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

<=> (3x + 1)(3x - 2) + 15(3x + 1) = 2(2x + 1)(3x + 1) + 6x(3x + 1)

<=> 9x² - 6x + 3x - 2 + 45x + 15 = 12x³ + 4x + 6x + 2 + 18x² + 6x

<=> 9x² + 42x + 13 = 30x² + 16x + 2

<=> 9x² + 42x + 13 - 30x² - 16x - 2 = 0

<=> -21x² + 26x + 11 = 0

<=> 21x² - 26x - 11 = 0

<=> 21x² + 7x - 33x - 11 = 0

<=> 7x(3x + 1) - 11(3x + 1) = 0

<=> (7x - 11)(3x + 1) = 0

<=> 7x - 11 = 0 hoặc 3x + 1 = 0

<=> x = 11/7 hoặc x = -1/3